代码随想录算法训练营第四十四天 | 动态规划 part11

1143.最长公共子序列

class Solution {
public:
    int longestCommonSubsequence(string text1, string text2) {
        int t1 = text1.size(), t2 = text2.size();
        vector<vector<int>> dp(t1 + 1, vector<int>(t2 + 1, 0));

        for (int i = 1; i <= t1; ++i) {
            for (int j = 1; j <= t2; ++j) {
                if (text1[i - 1] == text2[j - 1])
                    dp[i][j] = dp[i - 1][j - 1] + 1;
                else 
                    dp[i][j] = max(dp[i - 1][j], dp[i][j - 1]);
            }
        }
        return dp[t1][t2];
    }
};
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1035.不相交的线

class Solution {
public:
    int maxUncrossedLines(vector<int>& nums1, vector<int>& nums2) {
        int t1 = nums1.size(), t2 = nums2.size();
        vector<vector<int>> dp(t1 + 1, vector<int>(t2 + 1, 0));

        for (int i = 1; i <= t1; ++i) {
            for (int j = 1; j <= t2; ++j) {
                if (nums1[i - 1] == nums2[j - 1])
                    dp[i][j] = dp[i - 1][j - 1] + 1;
                else 
                    dp[i][j] = max(dp[i - 1][j], dp[i][j - 1]);
            }
        }
        return dp[t1][t2];
    }
};
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53. 最大子序和

class Solution {
public:
    int maxSubArray(vector<int>& nums) {
        // dp[i]表示以已nums[i]为结尾的最大子数组和
        // dp[i] = max(dp[i - 1] + nums[i], nums[i])
        // 初始化为0, dp[0] = nums[0]
        vector<int> dp(nums.size(), 0);
        dp[0] = nums[0];
        int result = nums[0];
        for (int i = 1; i < nums.size(); ++i) {
            dp[i] = max(dp[i - 1] + nums[i], nums[i]);
            if (dp[i] > result) result = dp[i];
        }
        return result;
    }
};
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392.判断子序列

class Solution {
public:
    bool isSubsequence(string s, string t) {
        int n1 = s.size(), n2 = t.size();
        vector<vector<int>> dp(n1 + 1, vector<int>(n2 + 1, 0));

        for (int i = 1; i <= n1; ++i) {
            for (int j = 1; j <= n2; ++j) {
                if (s[i - 1] == t[j - 1])
                    dp[i][j] = dp[i - 1][ j - 1] + 1;
                else 
                    dp[i][j] = dp[i][j - 1];
            }
        }
        return dp[n1][n2] == n1;
    }
};
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